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It follows from Zorn's Lemma that every vector space $V$ has a basis (this means, a subset $B$ of $V$ that generate any $v \in V$ by means a finite linear combination, and such that $B$ is LI) . But, seems quite impossible to me to find a basis for $\mathbb{R}^{\infty}$; I have listened somewhere that is really impossible find an actual basis for this space, because of something that I don't remember.

There exist a concrete basis? There exists some reason for we never shall find one?

Brian Fitzpatrick
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João Dos Reis
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    I'm not sure if this is universal notation, but from what I've seen $\mathbb{R}^{\infty}$ is a countable direct sum of $\mathbb{R}$'s and has a basis that is easy to describe (all but finitely many of the coordinates are $0$), while $\mathbb R^{\omega}$ is the one with the elusive uncountable basis. – Matt Samuel Jun 26 '15 at 21:46
  • @MattSamuel I was calling $\mathbb{R}^\infty$ the countable direct sum; but I don't think that this set is a base, since it's not linear independent $(1,1,0,0,0,...) - (1,0,0,0,0,...) - (0,1,0,0,0,..) = (0,0,0,0,..) $. – João Dos Reis Jun 26 '15 at 22:14

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Finding a basis for $\Bbb R^\infty$ depends on how you choose to define $\Bbb R^\infty$. Let's expand on @MattSamuel's comment by introducing some terminology.

The support of a function $f:X\to\Bbb R$ is the set $$ \DeclareMathOperator{supp}{supp}\supp(f)=\{x\in X:f(x)\neq0\} $$

One reasonable definition of $\Bbb R^\infty$ is $$ \Bbb R^\infty=\left\{\Bbb N\xrightarrow{f}\Bbb R:\lvert\supp(f)\rvert<\infty\right\} $$ It is not difficult that this definition makes $\Bbb R^\infty$ an $\Bbb R$-vector space. To exhibit a basis, for $j\in\Bbb N$ let $\chi_j:\Bbb N\to\Bbb R$ be $$ \chi_j(n)= \begin{cases} 1 & n=j \\ 0 & n\neq j \end{cases} $$ One then shows that $$ \beta=\{\chi_j:j\in\Bbb N\} $$ is a basis for $\Bbb R^\infty$.

If, however, one chooses to define $\Bbb R^\infty$ as the collection of all maps $f:\Bbb N\to\Bbb R$, then $\Bbb R^\infty$ is still an $\Bbb R$-vector space and therefore Zorn's lemma implies that $\Bbb R^\infty$ has a basis. This time a basis is much more difficult (or maybe even impossible!) to write down.

Brian Fitzpatrick
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